令x=√(a+5),∴x²=a+5
y=√(b+5),∴y²=b+5
Z=√(c+5),∴Z²=c+5
x²+y²+z²=a+b+c+15=16 ①
∵(x-y)²+(y-z)²+(z-x)²≥0
即2(x²+y²+z²)-2(xy+yz+xz)≥0
∴x²+y²+z²≥xy+yz+xz ②
由(x+y+z)²=(x²+y²+z²)+2(xy+yz+xz)≤3(x²+y²+z²)
∴(x+y+z)²≤48
得x+y+z≤4√3,
原不等式得证.
令x=√(a+5),∴x²=a+5
y=√(b+5),∴y²=b+5
Z=√(c+5),∴Z²=c+5
x²+y²+z²=a+b+c+15=16 ①
∵(x-y)²+(y-z)²+(z-x)²≥0
即2(x²+y²+z²)-2(xy+yz+xz)≥0
∴x²+y²+z²≥xy+yz+xz ②
由(x+y+z)²=(x²+y²+z²)+2(xy+yz+xz)≤3(x²+y²+z²)
∴(x+y+z)²≤48
得x+y+z≤4√3,
原不等式得证.